Question 16.3: Rotating Blade Design Analysis A disk of uniform thickness i...
Rotating Blade Design Analysis
A disk of uniform thickness is used at 12,000 rpm as a rotating blade for cutting blocks of paper or thin plywood. The disk is mounted on a shaft of 1 in. radius and clamped, as shown in Figure l6.9. Determine
a. The factor of safety n according to the maximum shear stress criterion
b. The values of the maximum radial stress and displacement at outer edge
Assumptions: The cutting forces are relatively small, and speed is steady: loading is considered static. The disk outside radius is taken as 15 in. The stress concentrations due to clamping and sharpening at the periphery are disregarded.
Design Decision: The disk material is a high-strength ASTM A242 steel.
The material properties are (Table B.1)
\begin{array}{c} \rho=\frac{0.284}{386}=7.358 \times 10^{-4} lb \cdot s ^2 / in ^4, \quad \nu=0.3 \\ E=29 \times 10^6 psi , \quad S_{y s}=30 ksi \end{array}
We have
\rho \omega^2=7.358 \times 10^{-4}\left(\frac{12,000 \times 2 \pi}{60}\right)^2=1161.929
a. The tangential stress, expressed by Equation 16.31b with p = 0, has the form
\sigma_\theta=\frac{3+\nu}{8}\left(a^2+b^2-\frac{1+3 v}{3+v} r^2+\frac{a^2 b^2}{r^2}\right) \rho \omega^2
The stresses in the inner and outer edges of the blade are, from the preceding equation,
\left(\sigma_\theta\right)_{r=1}=\frac{3.3}{8}\left\lgroup 1^2+5^2-\frac{1.9 \times 1^2}{3.3}+\frac{1^2 \times 5^2}{1^2} \right\rgroup (1161.929)=24.17 ksi
\left(\sigma_\theta\right)_{r=5}=\frac{3.3}{8}\left\lgroup 1^2+5^2-\frac{1.9 \times 5^2}{3.3}+1^2 \right\rgroup (1161.929)=6.04 ksi
The maximum shear stress occurs at the inner surface (r = 1 in.), where \sigma _{r} = 0:
\tau_{\max }=\frac{\sigma_\theta}{2}=\frac{24.17}{2}=12.09 ksi
The factor of safety, based on the maximum shear stress theory, is then
n=\frac{S_{y s}}{\tau_{\max }}=\frac{30}{12.09}=2.48
Comment: Should there be starts and stops, the condition is one of fatigue failure, and a lower value of n would be obtained by the techniques of Section 7.11.
b. The largest radial stress in the disk, from Equation 16.33, is given by
\begin{aligned} \sigma_{r, \max } &=\frac{3+ \nu }{8}(b-a)^2 \rho \omega^2 \\ &=\frac{3.3}{8}(5-1)^2(1161.929)=7669 ksi \end{aligned}
The radial displacement of the disk is expressed by Equation 16.31c with p = 0. Hence,
u=(u)_p+\frac{(3+\nu)(1-\nu)}{8 E}\left\lgroup a^2+b^2-\frac{1+v}{3+\nu} r^2+\frac{1+\nu}{1-\nu} \frac{a^2 b^2}{r^2} \right\rgroup \rho \omega^2 r (16.31c)
\begin{aligned} (u)_{r=5} &=\frac{(3.3)(0.7)}{8 \times 29 \times 10^6}\left\lgroup 1^2+5^2-\frac{1.3 \times 1^2}{3.3}+\frac{1.3 \times 1^2}{0.7} \right\rgroup (1161.929)(5) \\ &=1.589 \times 10^{-3} in . \end{aligned}
is the radial displacement at the outer periphery.